No prob... just please take time to read the posting guidelines which you had to agree to in order to register here. Also note the huge 'stickies' at the top of the sub-forums that say "Don't Post Your Homework Questions Here".
Again, though, you have to provide a lot more information before an answer can be arrived at. What have you done about figuring it out so far?

Staff: Mentor

Why do we need any variables.
According to the zeroth law of thermodynamics heat will be transferred from the surrounding to the system. Until the two that is the surrondings and the system reach equilibrium.

Why is there any need for variables to be supplied. I dont know. Please could you correct me if i am wrong.

Why do we need any variables.
According to the zeroth law of thermodynamics heat will be transferred from the surrounding to the system. Until the two that is the surrondings and the system reach equilibrium.

Why is there any need for variables to be supplied. I dont know. Please could you correct me if i am wrong.

oh...,my original mean focuses on the quantity of heat between the icebox to surroundings and the icebox self-generated by cooling the inner space of it. which is larger between this two quantity of heat? and im under the assumption that the fridge is a normal, powered fridge, that gets opened in standard room-temperature conditions.i think it should be refer to the 1st and 2nd laws of thermodynamics

oh...,my original mean focuses on the quantity of heat between the icebox to surroundings and the icebox self-generated by cooling the inner space of it. which is larger between this two quantity of heat? and im under the assumption that the fridge is a normal, powered fridge, that gets opened in standard room-temperature conditions.i think it should be refer to the 1st and 2nd laws of thermodynamics

Maybe you can tell us what you're trying to do. Is this a homework problem? What's the Author and Title of the book and maybe I can figure out what you're trying to ask better.

this problem is based on the naive problem "if i open an icebox's door, can i cool down the temperature of my house?"

maybe i can illustrate this problem in this way:i only care about two factors in this problem,one is that the inner space of the icebox can transport heat to the surroundings so as to lower the temperature of surroundings,the other is that the icebox would generate heat due to lowering the temperature of the inner space.you can feel the heat on the side surface of the icebox, right?and this heat can heighten the temperature of surroundings.so eventually what is the temperature of the surroundings?does it a little higher or lower than the original one?

assumption:the fridge is a normal, powered fridge that gets opened in standard room-temperature conditions,and the room is isolated.

this problem is based on the naive problem "if i open an icebox's door, can i cool down the temperature of my house?"

In that case, I'd say no. Not the overall average temperature of your house... it will just be transporting heat. Taking it from inside the fridge, and pumping it out the bottom or back of the fridge. Both of these points are inside your house, so the heat isn't going anywhere.

maybe i can illustrate this problem in this way:i only care about two factors in this problem,one is that the inner space of the icebox can transport heat to the surroundings so as to lower the temperature of surroundings,the other is that the icebox would generate heat due to lowering the temperature of the inner space.you can feel the heat on the side surface of the icebox, right?and this heat can heighten the temperature of surroundings.so eventually what is the temperature of the surroundings?does it a little higher or lower than the original one?

I imagine it would stay the same temperature in the house overall (given that the house is a sealed system), only the local temperatures at each radiator will have a local hot and a local cold temperature.

On the other hand, the cord that's powering the fridge's heat pump is suffering losses from entropy, so it's bringing energy in from the power station, so the heat could rise slightly, but that's probably nitpicking.

Staff: Mentor

To put together some things people said:

When you first open the door, the fridge will warm and the house will cool. Then the fridge compressor will turn on to try to maintain the fridge temperature and the house will warm, quickly passing its starting temperature. The fridge uses electricity - all of that electricity is converted to heat and warms up the house.

Assumptions are helpful, as long as you declare them and remain aware they are assumptions. This is how theoretical science develops. It's ok to be wrong as long as you can acknowledge it.

I cautiously make a great number of assumptions and follow hunches in most everyday situations; there isn't enough time in the day to analyze everything. On this site, however, that has backfired on me a few times. I've made perfectly logical, obvious assumptions about something, only to find out after wasting hours upon hours that the OP had something totally different in mind. Usually, it's a communications problem based upon either the age of the correspondent or a lack of English fluency, but sometimes it was just laziness on my part. Nowadays I just find it prudent to be sure of the facts first.

You initially said "you open an icebox's door". An icebox is not a refrigerator! An icebox is an insulated box containing ice and room for things you want to keep cool! Opening the door will cool the outside room and warm the inside of the icebox.

If you really meant a refrigerator, then I would make the reasonable assumptions that the refrigerator had been running for some time and the inside was cooler than the outside. Just opening the door will, again, reduce the temperature of the outside room slightly and increase the temperature inside.

A varient problem is this: take a refrigerator that is already at room temperature into a room, plug it in and turn it on (with the door open). what will happen to the temperature in the room?